What is the integral of sin(7x)cos(3x) ?

The integral of sin(7x)cos(3x) is 1/2 (-1/10 cos(10x)-1/4 cos(4x))+C

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integral of sin(7x)cos(3x)

\int sin (7 x) cos (3 x) d x

\frac{1}{2} (- \frac{1}{10} cos (10 x) - \frac{1}{4} cos (4 x)) + C

Solution steps

\int sin (7 x) cos (3 x) d x

Use the following identity:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

= \int \frac{sin ( 7 x + 3 x ) + sin ( 7 x - 3 x )}{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int sin (7 x + 3 x) + sin (7 x - 3 x) d x

Simplify

= \frac{1}{2} \cdot \int sin (10 x) + sin (4 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int sin (10 x) d x + \int sin (4 x) d x)

\int sin (10 x) d x = - \frac{1}{10} cos (10 x)

\int sin (4 x) d x = - \frac{1}{4} cos (4 x)

= \frac{1}{2} (- \frac{1}{10} cos (10 x) - \frac{1}{4} cos (4 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{10} cos (10 x) - \frac{1}{4} cos (4 x)) + C

\frac{d y}{d x} = 0.5 (y - 200)

\frac{d}{d x} (tanh (x))

x \to 2 lim (\frac{∣ x - 2 ∣}{x})

\frac{d}{d x} (\frac{x + 1}{x ^{2} + 3})

\int \frac{1}{( x ^{3} + 8 )} d x

What is the integral of sin(7x)cos(3x) ?
The integral of sin(7x)cos(3x) is 1/2 (-1/10 cos(10x)-1/4 cos(4x))+C